Standardization of NaOH Lab Report Part 2 PDF

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Bryahna Henriquez PID: 6241339 Due 25 October 2021 Section U TA Isabel Diaz Standardization of NaOH Complete Analysis ReportPurpose: The purpose of this lab was to perform uniform acid-base titrations, while standardizing an NaOH solution in a buret correctly. Also, the students were able to calcula...

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Bryahna Henriquez PID: 6241339 Due 25 October 2021 Section U04 TA Isabel Diaz Standardization of NaOH Complete Analysis Report

Purpose: The purpose of this lab was to perform uniform acid-base titrations, while standardizing an NaOH solution in a buret correctly. Also, the students were able to calculate molarity using stoichiometric calculations, and calculate the RAD for the data collected. Introduction: Acid-base reactions are often tied together with analytical analyses of various products. These reactions proceed essentially to completion, and are linked with fundamental stoichiometry. Acid-base reactions occur fast, reaching an equivalence point. This point is known as the point at which a reaction is complete or ends. In other terms, the equivalence point was the point at which the amount of titrant added was enough to fully neutralize the analyte solution. At this point, moles of base will equal moles of acid (Libretexts, 13.5: Acid/base titration). On the other hand, there are endpoints. This point is where the color change of the indicator is seen (Et. al, 2015). The indicator students used in this experiment was phenolphthalein, and it changes from clear in acids to pink in bases. This indicator was chosen to be used because it demonstrates the equivalence points anywhere between 8-10 on the pH scale. The titration involved a weak acid with a strong base, therefore the equivalence point will be greater than 7. Students performed volumetric analysis in this lab, this process involved measuring the known volume of a standard solution. A standard solution is often known as a solution for which the concentration is accurately known, and usually expressed in molarity and to four significant figures. The concentration expressed in molarity, M, equals moles of solute per liter of solution (Graves, 2008). The primary standard students used in this lab was a solid, with a known molecular weight; oxalic acid. A primary standard is a reagent used to dispense an accurately known

amount of analyte (Libretexts, 5.1: Analytical signals) ). This reagent is a stable compound known to have an extremely high purity (Anne Marie Helmenstine, 2020 . Standardization is a process used to determine the exact concentration of a prepared solution (Openstax, 2016). A standard solution can be prepared indirectly or directly. To be prepared directly, a certain mass of a primary standard is dissolved in a volumetric flask with a given volume of solution. To be prepared indirectly, a solution of approximately known concentration is prepared and standardized against a weighted amount of primary standard. Both ways require the concentration of the solution to be calculated from the weighted amount of primary standard and its molecular weight. Students used the indirect method to standardize the solid sodium hydroxide. The reaction scheme of the standardization NaOH with the oxalic acid is as follows: H 2 C 2 O 4 ❑( aq ) +2 NaOH( aq) → Na2 C2 O 4 +4 H 2 O In the lab, students will slowly add sodium hydroxide to the oxalic acid from a buret. The volume of sodium hydroxide used to reach the equivalence point is accurately known, allowing for the concentration of the solution to be calculated from the relationship of the values of these equations: mol NaOH=g ox acid ∗ molarity (NaOH )=

mol ox acid mol NaOH ∗ mol ox acid g ox acid

mol NaOH L NaOH

Students were able to calculate how many moles of NaOH were required for 1 mole of oxalic acid in the mole to mole ratio. To ensure the molarity of the sodium hydroxide solution, triplicate samples were titrated. When the triplicate samples obtain an average value that has a precision of 20 ppt or less relative average deviation, then the solution is standardized. Over 20 ppt indicates more trials need to be done. Once the standard of NaOH is reached, it can be used

as a secondary standard to tirate various other acid solutions or perform analyses. Diluting is known as the process of adding additional solvent to a solution to decrease its concentration (Mott, 2021). The dilution factor refers to the ratio of the volume of the initial (concentrated) solution to the volume of the final (diluted) solution. The dilution factor can be calculated with the equation (Stubbings, 2021): DF=

V2 V1

The relative average deviation (RAD) that was calculated by the experimenters for each part of the experiment is a percentage that portrays how much, on average, each measurement collected differers from the initial qualitative mean of data. This percentage is important because it allows the experimenters to determine how wide or narrow a curve plotted from the data points would be (Deziel, 2020). In the lab, the experimenters will be slowly adding sodium hydroxide to hydrochloric acid from a buret. The volume of sodium hydroxide used to reach the equivalence point is accurately known, allowing for the determination of the concentration of the unknown HCl solution that was used to be calculated. The reaction scheme for the standardization of NaOH with the unknown D HCl solution is as follows: HCl+ NaOH → NaCl + H 2 O

To ensure the molarity of the Unknown D HCl solution, triplicate samples were titrated. When the triplicate samples obtain an average value that has a precision of 5 ppt or less relative average deviation, then the solution is standardized. Over 5 ppt indicates more trials need to be done. The experimenters will also be proceeding with the same procedure but with acetic acid to titrate. The experimenters will slowly add sodium hydroxide to acetic acid from a buret. The students

will be determining the percent acetic acid in commercial vinegar. Diluted vinegar will be titrated against NaOH. The reaction scheme for the substances used is as follows: CH 3 COOH + NaOH → CH 3 COONa+ H 2 O Procedures: The experimenters obtained two volumetric flasks; a 250 mL and a 100 mL flask, a 10 mL pipette, a magnetic stir plate and rod, as well as a buret. Part 1 of the experiment began with the students preparing the stock solution with 17 mL 6 M NaOH, diluted with 983 mL of DI water. Three 250 mL erlenmeyer flasks were obtained and prepared with; 0.221 g of oxalic acid, 15 mL DI water, and 5 drops of the phenolphthalein indicator. The experimenters then prepared the buret, filling it to the 0.00 mL indication. Each flask was labeled trial 1-3. Flask 1 was obtained. The NaOH stock solution in the buret was slowly released from the buret into the flask until an equivalence point was reached. This point was indicated by a faint pink color that remains without going away. The measurement of the NaOH diluted with DI water was recorded in the data table. These procedures were repeated for flasks 2 and 3. The average molarity, mean, deviation from mean, and RAD were calculated, based on the results obtained from these flasks. When the triplicate samples obtain an average value that has a precision of 20 ppt or less relative average deviation, then the solution is standardized. Over 20 ppt indicates more trials need to be done. The percent yield was also calculated. The NaOH solution prepared was kept for parts 2 and 3 of the lab. For part 2 of the experiment, the students were assigned an unknown HCl. Unknown D was assigned to this group of students. To begin, 10 mL of the HCl unknown D was obtained, and put into a 100 mL beaker. The beaker was filled to the 100 mL mark with the NaOH stock solution. After diluting, a 10 mL pipette was used to transfer 10 mL of the diluted HCl unknown D

solution into three 250 erlenmeyer flasks. The three flasked were prepared with 10 mL of the solution and 5 drops of the phenolphthalein indicator. To titrate the triplicate samples, the same procedures done in part 1 were used. Flask 1 was obtained. The NaOH stock solution in the buret was slowly released from the buret into the flask until an equivalence point was reached. This point was indicated by a faint pink color that remains without going away. Observations and calculations were inputted into the data table. These procedures were repeated for flasks 2 and 3. The average molarity, mean, deviation from mean, and RAD were calculated, based on the results obtained from these flasks. When the triplicate samples obtain an average value that has a precision of 5 ppt or less relative average deviation, then the solution is standardized. Over 5 ppt indicated more trials needed to be done. For part 3 of the lab experiment, the students pipetted 10 mL of commercial vinegar into a 100 mL volumetric flask, and diluted to the 100 mL mark with DI water. The mixture was shaken well. The triplicate samples were titrated against the standard NaOH solution, with 5 drops of the phenolphthalein indicator being added. The same process used for titration in parts 1 and 2 were repeated for part 3. The average molarity of the diluted vinegar solution, as well as the molarity of the undiluted sample were calculated. The lab students were also able to calculate the dilution factor based on the results.

Results: Part 1 Results: Data Table 1: Recorded Mass of H2C2O4 (g) in Flask Flask #1

Flask #2

Flask #3

Flask #4

Flask #5

Mass weighing bottle before (g)

…………….... …………….... ……………… …………….... ……………....

Mass weighing bottle after (g)

……………… ……………… …………….... ……………… ……………....

Mass H2C2O4 in flask (g)

0.221 g

0.221 g

0.221 g

0.221 g

0.221 g

** 15 mL DI water used in each flask Data Table 2: Recorded Volumes of NaOH used and Molarity Flask #1

Flask #2

Flask #3

Flask #4

Flask #5

Initial volume in Buret (ml)

0.00 mL

0.00 mL

0.00 mL

0.00 mL

0.00 mL

Final Volume in Buret (ml)

66.50 mL

39.93 mL

36.40 mL

33.00 mL

29.50 mL

Volume NaOH used (ml)

66.50 mL

39.93 mL

36.40 mL

33.00 mL

29.50 mL

Molarity of NaOH (M)

0.0527 M

0.08778 M

0.0962 M

0.1062 M

0.11881

Average molarity (M)

0.07521 M

……………... …………….... …………….... ……………....

**Flask 4 added **Flask 1 & 2 will not be included for calculations for standard deviation (outlier).

**Flask 5 added Data Table 3: RAD Calculation Flask number

M NaOH

Deviation from Mean M NaOH – Mean

1

0.0527 M

OUTLIER (NOT USED)

2

0.08778 M

OUTLIER (NOT USED)

3

0.0962 M

0.02099

4(if needed)

0.1062 M

0.06459

5(if needed)

0.11881 M

0.0436

Mean

0.07521 M

0.04306

Calculations: H 2 C 2 O 4 ❑( aq ) +2 NaOH → Na2 C 2 O 4 + 4 H 2 O (6 M )( x )=(0.1 M )(1000 mL) x=16.67 ≈ 17 mL6 M NaOH +983 DI Water

35 mL ×

1L 0.1 M NaOH × 1mol H 2 C 2 O 4 126.1 g =0.221 g × × 1L 1mol 1000 mL 2mol NaOH

1) Volume NaOH: 66.50 mL−0.00 mL=66.50 2) Mol NaOH: 1 mol ox acid 2 mol NaOH × 1 ox acid 126.1 g ox acid −3 ¿ 3.5051 × 10 mol NaOH ¿ 0.003051 mol NaOH 3) Molarity: 0.221 g ox acid ×

Flask 1:

0.0035051mol NaOH =0.0527 M NaOH 0.0665 L NaOH

Flask 2:

0.0035051mol NaOH =0.08778 M NaOH 0.03993 L NaOH

Flask 3:

0.0035051mol NaOH =0.0962 M NaOH 0.0364 L NaOH

Flask 4:

0.0035051mol NaOH =0.01062 M NaOH 0.033 L

Flask 5:

0.0035051mol NaOH =0.11881 M NaOH 0.0295

4) Trial 1 (before flasks 4 &5 were added): Mean:

0.0527 + 0.08778 + 0.0962 =0.07889 3

Deviation: 1: 0.0527 −0.07889=0.02619 2: 0.08778− 0.07889=0.00981 3: 0.0962−0.07889 =0.01731 Average Deviation:

0.02619+ 0.00981+ 0.01731 =0.01800 3

RAD Calculation Trial 1: 0.01800 ×1000 =228.16 0.07889 Trial 2 (flask 4 added): Mean:

0.0527 + 0.08778 + 0.0962 + 0.01062 =0.06185 4

Deviation: 1: 0.0527 −0.06185=0.00915 2: 0.08778− 0.06185=0.02593 3: 0.0962−0.06185 =0.03435 4: 0.01062−0.06185 =0.05123 Average Deviation:

0.00915+ 0.02593 + 0.03435 + 0.05123 =0.030165 4

RAD Calculation Trial 2: 0.030165 ×1000 =487 0.06185 Rad Calculation for flasks 2, 3, 4:

Mean:

0.08778+ 0.01062+ 0.0962 =0.0648 3

Deviation: 1: 0.08778− 0.0648=0.02298 2: 0.0962−0.0648 =0.0314 3: 0.01062−0.0648 =0.05418 Average Deviation:

0.02298+ 0.0314 + 0.05418 =0.03618 3

RAD Calculation: 0.03618 ×1000 =558.3 0.0648

Trial 3 (flask 5 added): **Calculations only include flasks 3, 4, 5*** Mean:

0.0962+ 0.11881 + 0.01062 =0.07521 3

Deviation: 1: 0.0962−0.07521 =0.02099 2: 0.01062−0.07521 =0.06459 3: 0.11881 −0.07521=0.0436 Average Deviation:

0.02099+ 0.06459 + 0.0436 =0.04306 3

RAD Calculation: 0.04306 ×1000 =572 0.07521 5) % error 0.07521 M−0.1 M ×100 0.1 M ¿ 24.79 % Part 2 Results:

****UNKNOWN D**** ***10 mL used***** Data Table 4: Recorded Volume of NaOH used and Molarity of Unknown D HCl

Initial volume in Buret (ml)

Flask #1

Flask #2

Flask #3

00.00 mL

00.00 mL

00.00 mL

Final Volume in Buret (ml)

42.30 mL

42.50 mL

42.30 mL

Volume NaOH used (ml)

42.30 mL

42.50 mL

42.30 mL

Molarity of Unknown D (M)

0.3180 M HCl

0.3196 M HCl

0.318096 M HCl

Average molarity (M)

0.3185 M HCl

……………………... ……………………...

Data Table 5: RAD Calculation Flask number

M Unknown D

Deviation from Mean M Unknown D – Mean

1

0.3180 M

.0005

2

0.3196 M

.00011

3

0.318096 M

.0005

4(if needed)

…………………………….... ……………………………....

5(if needed)

…………………………….... ……………………………....

Mean

0.3185 M

Calculations:

1) Volume of NaOH: 42.30 mL −0.00 mL =42.30

2) HCl Molarity: Flask 1: Flask 2:

(0.0752 NaOH M )(0.0423 L NaOH ) =0.3180 M HCl 0.01acid (0.0752 NaOH M )(0.04250 L NaOH ) =0.3196 M HC l 0.01acid

.00037

Flask 3:

(0.0752 NaOH M )(0.04230 L NaOH ) =0.318096 M HCl 0.01acid

3) Average HCl Molarity: (0.3180 )+(0.3196 )+(0.31809) =0.3185 M 3

4) Deviation from Mean: Flask 1:

0.3180− 0.3185=.0005

Flask 2:

0.3196 −0.3185=.00011

Flask 3:

0.3180− 0.3185=.0005

5) Average Deviation: (.0005 )+(.0005 )+(0.00011 ) =.00037 3

6) RAD Calculation: .00037 ×1000 0.3185 ¿ 1.16

7) % error (unknown D): 0.3185 M − 0.5638 M × 100 0.5638 M ¿ 43.5 % Part 3 Results:

***10 mL diluted vinegar used*** Flask #1

Flask #2

Flask #3

Initial volume in Buret (ml)

00.00 mL

00.00 mL

00.00 mL

Final Volume in Buret (ml)

6.40 mL

5.70 mL

6.20 mL

Volume NaOH used (ml)

6.40 mL

5.70 mL

6.20 mL

Molarity of Vinegar (M)

0.0004827 M

0.000428 M

Average molarity (M)

0.00004589

……………………... ……………………...

Calculations:

1) Volume of NaOH: 6.40 mL−0.00 mL=6.40 mL

2) Acetic Acid Molarity: Flask 1: Flask 2: Flask 3:

(0.0752 )(0.00064) =0.0004827 M 0.01 (0.0752 )(0.000520) =0.000428 M 0.01 (0.0752 )(0.000620) =0.000466 M 0.01

3) Average Molarity: (0.0004827 )+( 0.000428)+(0.000466 ) =0.00004589 M 3

4) Undiluted Concentration: 0.00004589 ×10= 0.0004589

5) % error 0.00004589−0.05 × 100 0.05 ¿ 99.9 %

0.000466 M

Discussion/Conclusion: The results were tedious to obtain, and had an unstable change in NaOH. For part 1 of the lab, the students were able to calculate how many moles of NaOH were needed if the concentration was 0.1 M. Using this information, the students were able to calculate the molarity of each flask; the average molarity which was 0.07521; the average deviation which was 0.04306; and the RAD which was 572. While doing calculations, the students were not obtaining a low RAD calculation value, which led to the addition of two more flasks (flasks 4 & 5) to try and get the value lower. After all calculations were finished, the percent error was calculated. The percent error the students obtained was 24.79%. To obtain this value, the average molarity of flasks 3, 4, and 5 was used as the experimental value and 0.1 M as the theoretical value. While doing calculations, the students noticed the molarities of each flask were becoming bigger, creating outliers when calculating the mean. As seen in the data table 2, the students avoided using flasks 1 & 2 values to do calculations because they were outliers. This made the results vary, making the RAD larger rather than smaller. An issue encountered while titrating the flasks, was that the students overshot one of the flasks. Overshooting the flask, means that the color indicator was too pink. Too much of the solution was dropped too fast into the flask. Because the flask mixture was overshot, and too pink the values obtained could not be used, and the students had to continue to do more trials to obtain more accurate results. An error encountered was that we did not keep the solution in the bottle covered at all times. The TA had to tell the students to keep it closed at all times. This error introduces the likelihood of contamination via air particles, in turn skewing the measurements and calculations. This messes with the

concentrations calculated for the students’ results. The actual volume added to the flask was larger compared to the estimated value added to the flask. Flask 1 required 66.50 mL of base solution to reach

the equivalence point, while flask 3 only required 33.00 mL of base solution for the acid-base solution to reach the equivalence point. In part 2 of the lab experiment, the volume of the NaOH used to reach the equivalence point in titration with HCl were quite close. This allowed the students to determine the amount of NaOH solution needed to reach the HCl solution equivalence point. In this part of the experiment, the average molarity calculated was 0.3185; the average deviation calculated was 0.00037, and the RAD calculated was 1.16. This part also carried a % error of 43.5%. In part 3, the average molarity calculated for the diluted solution was 0.00004589 M, while the average molarity of the undiluted solution was 0.0004589 M. The undiluted concentration was found using a dilution factor, this allowed the lab students to be able to determine the undiluted concentration of acetic acid. This % error calculated for this part was very big, landing at 99.9%. Another error encountered in the duration of the experiment was in the transfer of the oxalic acid from the weighing paper. This affects the results of the experiment because some of the substance could have remained on the weighing paper, not fully being transferred off, which could skew the initial weight seen and recorded from the scale. This varies the amount of oxalic acid used for each of the triplicate samples, therefore having inaccurate, unreliable results. Yes, the students were able to successfully standardize the NaOH solution, but for part ...